**Example 1**. A child was asked to add first few natural numbers (i.e., 1 + 2 + 3 + ···) so long his patience permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the child discovered he had missed one number in the sequence during addition. Find the number he missed.

**Example 2**. In Δ*ABC*, *AB* = 4 cm, *BC* = 8 cm and *CA* = 6 cm. Find the length of the median drawn from *A* to *BC*.

**Example 3**. Prove that there is no positive integer in the interval *I* = (0, 1).

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “**the only way to learn mathematics is to do mathematics**”.

**Solution 1**. Note that, the boy gave the sum as 575.

Let us first try to see that summation till what number gives us a value close to 575.

Sum of first 10 natural numbers = 55.

Sum of first 20 natural numbers = 210.

Sum of first 30 natural numbers = 465.

Probably we need to add some more terms to get a sum close to 575.

Sum of first 31 natural numbers = 496.

Sum of first 32 natural numbers = 528.

Sum of first 33 natural numbers = 561.

But the sum of first 34 natural numbers = 595.

Hence, the missed out number is 595 – 575 = 20.

**Solution 2**. Let *D* be the mid-point of *BC*. By Apollonius theorem, if in a triangle the three sides are known, then the sum of the squares of two sides equal to twice the sum of squares of the median to the third side and the half of the third side.

Therefore, we can write

*AB*^{2}+*AC*^{2}= 2[*AD*^{2}+ (*BC*/2)^{2}] = 2[*AD*^{2}+*BD*^{2}]- or,
*AB*^{2}+*AC*^{2}= 2[*AD*^{2}+*BD*^{2}] - or, 4
^{2}+ 6^{2}= 2[*AD*^{2}+ 4^{2}] - or, 52 = 2[
*AD*^{2}+ 4^{2}] - or,
*AD*^{2}+ 16 = 26 - or,
*AD*^{2}= 26 – 16 = 10 - or,
*AD*= √10

Required length of the median is √10 cm.

**Solution 3**. If possible, assume that *I* is non-empty. Since *I* = (0, 1), it is the set of positive integers.

Being a set of positive integers, it must contain a least element, say *m*. Then 0 < *m* < 1.

Multiplying both sides of *m* < 1 by *m* gives *m*^{2} < *m*.

The square of a positive integer is a positive integer, so we have 0 < *m*^{2} < *m* < 1.

But this is saying that *I* has a positive integer *m*^{2} which is smaller than its least positive integer *m*.

This is a contradiction. Thus, we have *I* = *ϕ*. Hence the result.

It’s really wonderful way to learn mathematics. The way you explain mathematics can be hoon for the kids.

Keep going and helping students…

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Thank you so much sir

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